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# State which pairs of triangles in the given figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

"

To do:

We have to find whether the pair of triangles is similar and write the pair of similar triangles in the symbolic form in each case.

Solution:

(i) In $\triangle ABC$ and $\triangle PQR$,

$\angle A=\angle P=60^o$

$\angle B=\angle Q=80^o$

$\angle C=\angle R=40^o$

Therefore, by AAA criterion, we get,

$\triangle ABC \sim \triangle PQR$

(ii) In $\triangle ABC$ and $\triangle PQR$,

$\frac{BC}{PR}=\frac{2.5}{5}=\frac{1}{2}$

$\frac{AB}{QR}=\frac{2}{4}=\frac{1}{2}$

$\frac{AC}{PQ}=\frac{3}{6}=\frac{1}{2}$

Therefore, by SSS criterion, we get,

$\triangle ABC \sim \triangle QRP$

(iii) In $\triangle LMP$ and $\triangle EFD$,

$\frac{LM}{EF}=\frac{2.7}{5}=\frac{27}{50}$

$\frac{LP}{DF}=\frac{3}{6}=\frac{1}{2}$

$\frac{MP}{DE}=\frac{2}{4}=\frac{1}{2}$

$\frac{LM}{EF}≠\frac{LP}{DF}=\frac{MP}{DE}$

Therefore

$\triangle LMP$ is not similar to $\triangle EFD$

(iv) In $\triangle MNL$ and $\triangle PQR$,

$\frac{MN}{PQ}=\frac{2.5}{5}=\frac{1}{2}$

$\frac{ML}{QR}=\frac{5}{10}=\frac{1}{2}$

$\angle M=\angle Q=70^o$

Therefore, by SAS criterion, we get,

$\triangle NML \sim \triangle PQR$.

(v) In $\triangle ABC$ and $\triangle DEF$,

$\frac{AB}{DF}=\frac{2.5}{5}=\frac{1}{2}$

$\frac{BC}{EF}=\frac{3}{6}=\frac{1}{2}$

$\angle A=\angle F=80^o$

In $\triangle ABC, \angle A=80^o$ is not included between the sides.

Therefore,

$\triangle ABC$ is not similar to $\triangle DEF$.

(vi) In $\triangle DEF$,

$\angle F=180^o-(80^o+70^o)$

$=180^o-150^o$

$=30^o$

In $\triangle DEF$ and $\triangle PQR$,

$\angle E=\angle Q=80^o$

$\angle F=\angle R=30^o$

Therefore, by AA criterion, we get,

$\triangle DEF \sim \triangle PQR$.

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